Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  int n, m;
  cin >> n >> m;
  vector<vector<int>> v(n, vector<int>(m));
  int sum = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < m; j++) {
      cin >> v[i][j];
      sum += v[i][j];
    }
  }
  string col = "col", row = "row";
  if (m < n) {
    vector<vector<int>> u(m, vector<int>(n));
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        u[i][j] = v[j][i];
      }
    }
    swap(n, m);
    swap(col, row);
    v = u;
  }
  vector<int> ans1, ans2;
  for (int i = 0; i < n; i++) {
    int curmin = -1;
    for (int j = 0; j < m; j++) {
      if (curmin == -1 || curmin > v[i][j]) {
        curmin = v[i][j];
      }
    }
    if (curmin != 0) {
      for (int j = 0; j < curmin; j++) {
        ans1.push_back(i);
      }
    }
    for (int j = 0; j < m; j++) {
      v[i][j] -= curmin;
      sum -= curmin;
    }
  }
  for (int j = 0; j < m; j++) {
    int curmin = -1;
    for (int i = 0; i < n; i++) {
      if (curmin == -1 || curmin > v[i][j]) {
        curmin = v[i][j];
      }
    }
    if (curmin > 0) {
      for (int i = 0; i < curmin; i++) {
        ans2.push_back(j);
      }
    }
    for (int i = 0; i < n; i++) {
      v[i][j] -= curmin;
      sum -= curmin;
    }
  }
  if (sum != 0) {
    cout << -1 << endl;
    return 0;
  }
  cout << ans1.size() + ans2.size() << endl;
  for (int i = 0; i < ans1.size(); i++) {
    cout << row << ' ' << ans1[i] + 1 << endl;
  }
  for (int i = 0; i < ans2.size(); i++) {
    cout << col << ' ' << ans2[i] + 1 << endl;
  }
}